Problem 7 - Algebra and Trigonometry

Question

Use mathematical induction to prove that the formula is true for all natural numbers \(n\). $$ 1 \cdot 2+2 \cdot 3+3 \cdot 4+\cdots+n(n+1)=\frac{n(n+1)(n+2)}{3} $$

Step-by-Step Solution

Answer

The formula is true for all natural numbers by mathematical induction.

Step 1: Base Case

Start by proving the formula for the smallest natural number, which is usually 1. Substitute \(n=1\) into the formula to get the left-hand side sum and evaluate it:\[1 \cdot 2 = 2\]Next, substitute \(n=1\) into the right-hand side of the equation:\[\frac{1(1+1)(1+2)}{3} = \frac{1 \cdot 2 \cdot 3}{3} = 2\]Since the left-hand and right-hand sides are equal, the base case holds true.

Step 2: Inductive Hypothesis

Assume that the formula holds true for some arbitrary natural number \(k\), i.e.,\[1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \,\ldots\,+ k(k+1) = \frac{k(k+1)(k+2)}{3}\]This assumption is called the inductive hypothesis and will be used to prove the formula for the next natural number.

Step 3: Inductive Step

Now, prove the formula for \(n = k + 1\) using the inductive hypothesis:Consider the sum:\[1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \,\ldots\, + k(k+1) + (k+1)(k+2)\]According to the inductive hypothesis, the sum of the first \(k\) terms is \(\frac{k(k+1)(k+2)}{3}\), so we can write:\[\frac{k(k+1)(k+2)}{3} + (k+1)(k+2)\]This can be factored as:\[(k+1)(k+2) \left(\frac{k}{3} + 1\right) = (k+1)(k+2) \left(\frac{k+3}{3}\right)\]Simplify it:\[\frac{(k+1)(k+2)(k+3)}{3}\]This matches the right-hand side of the original formula for \(n = k+1\), thus proving the inductive step.

Step 4: Conclusion: Proof by Induction

Since the formula holds for the base case \(n=1\), and assuming it holds for an arbitrary \(k\) implies it also holds for \(k+1\), by mathematical induction, the formula is true for all natural numbers \(n\).

Key Concepts

Base Case

In mathematical induction, the **base case** is the first step to establishing that a given statement is true for all natural numbers. Generally, this means starting with the smallest natural number, typically 1.

For our exercise, we substitute 1 for the variable in the provided formula and evaluate both sides of the equation. On the left, we compute:

• \( 1 \cdot 2 = 2 \)

On the right side of the equation, we compute by substituting 1:

• \( \frac{1(1+1)(1+2)}{3} = \frac{1 \cdot 2 \cdot 3}{3} = 2 \)

Both sides equal 2, thus confirming the formula holds at this base level. Establishing this base confirms that the statement is accurate when applied to the smallest natural number.

Inductive Hypothesis

The **inductive hypothesis** is a crucial step in mathematical induction. This step involves assuming the statement is true for a random but specific natural number, say \(k\). This assumption forms the base for proving the next step.

In this exercise, we assumed the formula is true for \(n=k\):

• \( 1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \cdots + k(k+1) = \frac{k(k+1)(k+2)}{3} \)

By accepting this hypothesis as true, without yet proving it, we establish a foundation to prove that adding a next term in the sequence will also uphold the original formula's truth.

Inductive Step

In the **inductive step**, we aim to verify that if the formula holds for \(n=k\), then it must hold for \(n=k+1\). This step seeks to demonstrate the progression of truth along the sequence of natural numbers.

We start by expressing the sum up to \(n=k+1\) based on our inductive hypothesis:

• \( 1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \cdots + k(k+1) + (k+1)(k+2) \)
• Using the hypothesis, this can be rewritten as:\[\frac{k(k+1)(k+2)}{3} + (k+1)(k+2) \]
• Factor out the common term: \[(k+1)(k+2) \left(\frac{k}{3} + 1\right) = (k+1)(k+2) \frac{(k+3)}{3} \]
• Simplify to match the original formula for \(n=k+1\): \[\frac{(k+1)(k+2)(k+3)}{3} \]

By confirming that this match occurs, the induction step shows our formula is valid for \(n=k+1\), thereby proving the statement for all subsequent natural numbers.

Natural Numbers

The concept of **natural numbers** is paramount when discussing mathematical induction. Natural numbers are a set of positive integers beginning from 1 and increasing in unit increments: 1, 2, 3, and so on.

In our proof, induction is applied to these numbers to ensure that a formula is valid for every step within this set. The properties of natural numbers facilitate this incremental procedure by ensuring there is always a next number to move to—essential for linking each base step to its successor through the inductive process.

Induction leverages this consecutive nature by showing a base case works and then proving that, should a given natural number satisfy the condition, every next natural number does as well. This process covers an infinite set by planting these checks at key intervals along the series, providing a complete proof across the entire set of natural numbers.